MBA Questions of the day (27 Sep’13)

 Choose the appropriate alternative :

51. A company has a job to prepare certain number cans and there are three machines A, B & C for this job. A can complete the job in 3 days, B can complete the job in 4 days and C can complete the job in 6 days. How many days the company will take to complete the job if all the machines are used simultaneously?

(a) 4 days

(b) 4\3 days

(c) 3 days

(d) 12 days

52. n³ is odd. Which of the following statement(s) is(are) true.

I.   n is odd

II.  n² is odd

III. n² is even

(a) I only

(b) II only

(c) I and II only

(d) I and IIIonly

Q53 & 54: Production pattern for number of units (in cubic feet) per day.

For a truck that can carry 2000 cubic feet, hiring cost per day is Rs.1000. Storing cost per cubic feet is Rs.5 per day.

53. If all the units should be sent to the market, on which days should the trucks be hired to minimize the cost?

(a) 2^nd , 4^th , 6^th , 7^th

(b) 7^th

(c) 2^nd , 4^th , 5^th , 7^th

(d) None of these.

54. If storage cost reduced to Rs.0.8 per cubic feet per day, then

(a) 4^th

(b) 7^th

(c) 4^th and 7^th

(d) None of these

55. One bacteria splits into eight bacteria of the next generation. But due to environment only 50% of one generation can produce the next generation. If the seventh generation number is 4096 million, what is the number in first generation?

(a) 1 million

(b) 2 million

(c) 4 million

(d) 8 million

Answers:

51. In one day A would do 1/3^rd of the job, B would do 1/4^th of the job and C would do 1/6^th of the job. Hence if all three of them work simultaneously, in one day they would do (1/3 + ¼ +1/6) = 3/4^th of the job. Hence to complete the entire job together they would take 4/3 days.

52. If n³ is odd then n should also be odd. Hence n should also be odd. And n² will again be odd and not even. So only I and II are true.

53. We can see that there are two types of cost : (i) Transportation cost (viz. hiring truck) and (ii) storing cost. It can be seen that the daily production is far less than the capacity of the truck. So a truck can be hired to carry multiple days production at one go. So as long as the storing cost is less than the cost of hiring the truck (i.e. Rs.1000), it makes sense to store the production. When the storing cost exceeds Rs.1000, it is best that the entire lot be sent to the market. The cost pattern is as given in the following table:

* In spite of the fact that storing is cheaper than hiring truck on the last day, we have to with the latter option because everything that is manufacture has to sent to the market.

So according to this table if the truck is hired on 2^nd , 4^th , 6^th and 7^th day, total cost = Rs.5900. But is this the most cost effective scheme? It can be seen that we are hiring truck on two consecutive days (6^th & 7^th ). Hence since everything that is manufacture has to sent to the market, we have yet another option of hiring the truck on the 5^th day and sending the 6^th & 7^th days production together on the last day. In that case the cost on 5^th day would be Rs.1000 (i.e. Rs.200 more than the present cost), the cost on the 6^th day would be (120×5) = Rs.600 (i.e. Rs.400 less than the present cost) and the cost on the 7^th day would be Rs.1000 (The same as the present cost). Hence we can that the total cost would actually come down by (+200 – 400= – 200) Rs.200. Hence this becomes the most cost effective scheme. So we should hire trucks on 2^nd , 4^th , 5^th and 7^th day.

54. If the storage cost is reduced to Rs.0.8 per cubic feet then the cost pattern is as given in the following table :

* In spite of the fact that storing is cheaper than hiring truck on the last day, we have to with the latter option because everything that is manufacture has to sent to the market.

Hence the most cost effective scheme would be to send the entire production on the 7^th day.

55. Let us try and find a pattern. Let there be x bacteria in the first generation. Hence n₁ = x. But only x/2 among them will be able to produce the next generation. And they would give rise to 8(x/2) = 4x baxteria. Hence n₂ = 4x. Of these only 2x will give rise to next generation. So number of bacteria in the third generation = 8(2x) = 16x. So n₃ = 16x. Similarly we would find that n₄ = 64x. So if we observe : n₁ = x, n₂ = 4x, n₃ = 16x, n₄ = 64x. So this will form a GP with a = x and r = 4. The seventh term of this GP = 4096. So we can write, 4096 = x(4)⁶ = x(2)¹² = 4096x. Hence x = 1 i.e. 1 million.